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3 years ago

Here is to the people who need points (again) Have a good day!

Engineering

2 answers:

alexdok [17]3 years ago

8 0

Thank you so much!! You too!

g100num [7]3 years ago

7 0

Answer:

SERIOUSLY! TY

Explanation:

I don't think you understand how much this is gonna help me out :) Have an amazing summer! Stay safe!

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Steam enters an adiabatic nozzle at l MPa, 260 C, 30 m/s and exits at 0.3 MPa and 160 'C. Calculate the velocity at the exit.

maxonik [38]

Answer:

v₂ = 35.57 m/s

Explanation:

Given :

Inlet steam pressure, P₁ = 1 MPa

Inlet steam temperature, T₁ = 260°C

Inlet velocity of steam, V₁ = 30 m/s

Outlet steam pressure, P₂ = 0.3 MPa

Outlet steam temperature, T₂ = 160°C

Now,

From steam table at pressure 1 Mpa and temperature 260°C, enthalpy, h₁ = 2964.8 kJ/kg

From steam table at pressure 0.3 Mpa and temperature 160°C, enthalpy, h₂ = 2782.14 kJ/kg

Therefore, for an open system from 1st law of thermodynamics, we get

Energy in = Energy out

E₁ = E₂

$\left ( h_{1}+\frac{v_{1}^{2}}{2} \right )$ = $\left ( h_{2}+\frac{v_{2}^{2}}{2} \right )$

$v_{2}^{2}= 2\left [ h_{1}-h_{2}+\frac{v_{1}^{2}}{2} \right ]$

$v_{2}^{2}= 2\left [ 2964.8-2782.14+\frac{30^{2}}{2} \right ]$

$v_{2}^{2}=$ 2 X 632.66

v₂ = 35.57 m/s

Therefore, outlet velocity, v₂ = 35.57 m/s

7 0

3 years ago

1. Which of the following is the ideal way to apply pressure onto pedals?

Vikki [24]

I think D. By pressing gradually

8 0

3 years ago

Read 2 more answers

Air enters an adiabatic gas turbine at 1590 oF, 40 psia and leaves at 15 psia. The turbine efficiency is 80%, and the mass flow

melamori03 [73]

Answer:

a) 158.4 HP.

b) 1235.6 °F.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:

$m_{in}h_{in}=W_{out}+m_{out}h_{out}$

Whereas the mass flow is just the same, which means we have:

$W_{out}=m_{out}(h_{out}-h_{in})$

And the enthalpy and entropy of the inlet stream is obtained from steam tables:

$h_{in}=1860.7BTU/lbm\\\\s_{in}= 2.2096BTU/lbm-R$

Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:

$s_{out}= 2.2096BTU/lbm-R$

Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.

Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:

$h_{out}=1659.15BUT/lbm$

Then, the isentropic work (negative due to convention) is:

$W_{out}=2500lbm/h(1659.15BUT/lbm-1860.7BUT/lbm)\\\\W_{out}=-503,875BTU$

And the real produced work is:

$W_{real}=0.8*-503875BTU\\\\W_{real}=-403100BTU$

Finally, in horsepower:

$W_{real}=-403100BTU/hr*\frac{1HP}{2544.4336BTU/hr} \\\\W_{real}=158.4HP$

Regards!

6 0

3 years ago

A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °

SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)= $\frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3} (Btu/s)/(ft.F)$

Thermal Conductivity is SI units:

$k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K$

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

$A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J$

In Btu:

$A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu$

PArt B:

At midpoint Length=L/2=0.1 m

$Q=\frac{k*A*(T_1-T_2)}{L}$

On rearranging:

$T_2=T_1-\frac{Q*L}{KA}$

$T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C$

4 0

3 years ago

A particle travels along a straight line with a velocity v = (12 – 3t2) m/s. When t = 1 s, the particle is located 10 m to the l

arlik [135]

Answer:

The displacement from t = 0 to t = 10 s, is -880 m

Distance is 912 m

Explanation:

$v = (12 - 3t^2) m/s = ds/dt$ . . . . . . . . . . A

integrate above equation we get

$s = 12t - t^3 + C$

from information given in the question we have

t = 1 s, s = -10 m

so distance s will be

-10 = 12 - 1 + C,

C = -21

$s(t) = 12t - t^3 - 21$

we know that acceleration is given as

$a(t) = dv/dt = -6t$

[FROM EQUATION A]

Acceleration at t = 4 s, a(4) = -24 m/s^2

for the displacement from t = 0 to t = 10 s,

$s(10) - s(0) = (12*10 - 10^3 - 21) - (-21) = -880 m$

the distance the particle travels during this time period:

let v = 0,

$3t^2 = 12$

t = 2 s

Distance $= [s(2) - s(0)] + [s(2) - s(10)] = [1\times 2 - 2^3] + [(12\times 2 - 2^3) - (12\times 10 - 10^3)] = 912 m$

7 0

4 years ago

Here is to the people who need points (*again*) Have a good day!

Here is to the people who need points (again) Have a good day!